Advanced Aqueous Geochemistry
DM Sherman, University of Bristol
2010/2011
Oxidation-Reduction
Equilibria
Advanced Aqueous Geochemistry
DM Sherman,University of Bristol
Oxidation States of Atoms and Ions
• The oxidation state of an atom in an elemental form is 0.
In O2, O is in the 0 oxidation state. • When bonded to something else, oxygen is in oxidation
state -2 and hydrogen is in oxidation state of +1 (except for
peroxide and superoxide).
In CO32-, O is in -2 state, C is in +4 state.
• The oxidation state of a single-atom ion is the charge on
the ion.
For Fe2+, Fe is in +2 oxidation state.
1
Page 1
Advanced Aqueous Geochemistry
DM Sherman, University of Bristol
2010/2011
Oxidation-Reduction Reactions
Consider the reaction
2Fe2+ + MnO2 + 4H2O = 2Fe(OH)3 + 2H+ + Mn2+
• Fe2+ is being oxidized to Fe3+ (as Fe(OH)3)
• Mn4+ (as MnO2) is being reduced to Mn2+
We can express the overall reaction as two halfreactions:
2Fe2+ + 6H2O = 2Fe(OH)3 + 6H+ + 2eMnO2 + 4H+ + 2e- = Mn2+ + 2H2O
_________________________________
2Fe2+ + MnO2 + 4H2O = 2Fe(OH)3 + 2H+ + Mn2+
K for half-reactions
For each half-reaction,
A + e" = B
We can define an equilibrium constant
!
K=
[B]
[A][e " ]
Where [e-] is the activity of electrons. (This does not
mean that bare electrons are floating around in
solution!!)
! For convenience, take -log of the K
expression to get
pK = pB " pA " pe
!
2
Page 2
Advanced Aqueous Geochemistry
DM Sherman, University of Bristol
2010/2011
Important half reactions
Reaction
pK
1/4O2 + e- + H+ = 1/2H2O
20.75
2H+ + e- = H2
0
Cu+2 + e- = Cu+
-2.7
Fe3+ + e- = Fe2+
-13.02
MnO2 + 4 H+ + 2 e- = Mn+2 + 2 H2O
-41.38
CO3-2 + 10 H+ + 8 e- = CH4 + 3 H2O
-41.07
SO4-2 + 9 H+ + 8 e- = HS- + 4 H2O
-33.65
2 NO3- + 12 H+ + 10 e- = N2 + 6 H2O
-207.08
The pK ladder
3
Page 3
Advanced Aqueous Geochemistry
DM Sherman, University of Bristol
2010/2011
The pe concept..
By analogy with pH, the pe (-log[e-]) can be used to
characterize the redox state of a system.
We can define an equilibrium constant
K=
[B]
[A][e " ]
Where [e-] is the activity of electrons. (This does not
mean that
! bare electrons are floating around in
solution!!)
Range of pe of Aqueous Solutions
By convention, ΔG0 = 0.0 for the reaction
H+(aq) + e- = 1/2H2(g)
K=
(pH 2 )1/ 2
[H + ][e " ]
=1
The most reducing condition that is possible at the
Earth’s surface will have pH2 = 1 bar. Hence,
pH
!
+ pe = 0 or
pH = -pe
4
Page 4
Advanced Aqueous Geochemistry
DM Sherman, University of Bristol
2010/2011
Range of pe of Aqueous Solutions (cont.)
The most oxidizing condition under which an aqueous
solution can exist is buffered by the half-reaction
1/2H2O = e- +1/4O2 + H+
K=
(pO 2 )1/ 4 [H +][e"]
1/ 2
[H2O]
= 10 "20.75
Under the most oxidizing condition, pO2 =1. Since
[H2O] = 1, we have pH + pe = 20.75 or
!
pe = 20.75-pH
pe-pH Environments
5
Page 5
Advanced Aqueous Geochemistry
DM Sherman, University of Bristol
2010/2011
Redox Couples in Groundwater
pe-pH Diagrams: The Fe-H2O system
A
B
Fe3+ + e- = Fe2+
pe = -pK
Fe(OH)3 + 3H+ = Fe3+ + 3H2O
pH = (-pK + pFe3+)/3
Fe(OH)3 + 3H+ + e-
C
*[Fe]total = 10-3 m
6
= Fe2+ + 3H2O
pe = -pK- 3pH + pFe2+
Page 6
Advanced Aqueous Geochemistry
DM Sherman, University of Bristol
2010/2011
Calculating pe from Concentrations
Example: Given that pK = -16.5 for the half-reaction
Fe(OH)3(s) + 3H+ + e- = Fe2+ + H2O
calculate the pe of groundwater in which [Fe2+] = 10-4 M,
pH = 7 and the solution is saturated in Fe(OH)3.
Solution:
K=
[Fe 2+ ]
[H + ]3 [e " ]
pK = p[Fe] - 3pH - pe
!
-16.5 = 4 - 21 - pe Hence, pe = -0.5.
Predicting Stability of Species
Would (NO3)- be stable in this groundwater? pK = -104.6
for the reaction
NO3- + 6H+ + 5e- = 1/2N2(g) + 3H2O
K=
(PN2 (g) )1/2
-
[NO3 ][H+ ] 6[e " ]5
pK = -(1/2)log(PN2) -p[NO3-] - 6pH - 5pe
!
Since, PN2 = 0.8, pe = -0.5 and pH = 7, we get
-104.6 = -(1/2)(-0.097) -p[NO3-] - 42.0 +2.5
p[NO3-] = 65.1
7
Page 7
Advanced Aqueous Geochemistry
DM Sherman, University of Bristol
2010/2011
Voltage and Free Energy
Charge x Voltage = Energy
The free energy change (∆G) associated with the
transfer of n electrons is equivalent to a voltage drop
(E) experienced by the electrons:
∆G = -nFE
Where F is the faraday constant which is the charge
(in coulombs) of a mole of electrons (F = 9.64846 x
104 C/mol).
The Nernst Equation
By convention, we write half-reactions as reductions:
aA + bB + ne " # cC + dD
The voltage of this reaction is:
!
RT [A]a [B]b
E=E +
ln
nF [C ]c [D ]d
0
!
8
Page 8
Advanced Aqueous Geochemistry
DM Sherman, University of Bristol
2010/2011
The Nernst Equation (cont.)
At 25 ºC, we can write
0.059
[A]a [B]b
E=E +
log c d
n
[C ] [D ]
0
!
Be careful with the sign convention. Notice the
oxidized species are on top..
The Standard Hydrogen Electrode
We assign a voltage of 0 for the reaction
H + (aq ) + e " =
1
H2 (g)
2
when [H+] = 1 mole/liter and PH2(g) = 1 bar.
Hence,
Eo for the H+ reduction is 0.
!
Eh = "0.059log(PH2 )1/ 2 " 0.059pH
!
9
Page 9
Advanced Aqueous Geochemistry
DM Sherman, University of Bristol
2010/2011
The Standard Hydrogen Electrode
This cell would allow
us to measure the E
for the Cu2+-Cu half
reaction relative to
the S.H.E.
The salt bridge
allows ions, but not
electrons, to pass.
What defines the Eh of a System?
Ideally, in a system where
several of these redox
couples are present, they
would all be at equilibrium
and give the same Eh
value.
In practice, that is seldom the case. The Eh
measured with a Pt electrode is determined by the
most labile couple in the system.
10
Page 10
Advanced Aqueous Geochemistry
DM Sherman, University of Bristol
2010/2011
Dissolved Oxygen
• It is often difficult to get a reliable measurement of Eh (or
pe). An alternative is to measure dissolved oxygen.
The solubility of oxygen in water is given by Henry s law:
[O2] =KHPO2
With KH = 1.26 x 10-3 M/L-bar at 25 ºC
• Partial pressure of oxygen (O2) in atmosphere is 0.21
bar.
• Air-saturated water has a dissolved oxygen content of
2.6 x 10-4 mol O2/liter or 8.5 mg/L.
Berner’s Classification of Sediments
11
Oxic
O2 > 30mM
Fe2O3, FeOOH, MnO2
No organic matter.
Subo xic
30mM > O2
> 1 mM
Ano xic
O2 < 1 mM
Fe2O3, FeOOH, MnO2
Minor organic matter.
Sulfidic
H2S > 1 mM
FeS2, MnCO3, organic
matter
Nonsulfidic
H2S < 1 mM
FeCO3, Fe3(PO4)2,
MnCO3, low T Fe(II,III)
silicates
Page 11
Advanced Aqueous Geochemistry
DM Sherman, University of Bristol
2010/2011
Dissolved Oxygen and pe
The pe corresponding to a dissolve oxygen
concentration would be controlled by the reaction
(1/4)O2(aq) + e- + H+ = (1/2)H2O pK = 21.5
pe = 21.5 -pH + (1/4)log[O2(aq)]
In practice, the pe measured from Eh or derived from
other redox couples will differ greatly from that derived
from the dissolved O2.
Redox Couples in Phreeqc
SOLUTION 1 SEAWATER FROM NORDSTROM ET AL. (1979)
units
ppm
pH
8.22
pe
8.451
Default pe
density 1.023
temp
25.0
Couple which determines pe
redox
O(0)/O(-2)
Fe
0.002
Speciate using pe from O(0)
Zn
0.001
Mn
0.0002 pe
Use default pe
Al
0.0008
Si
4.28
Cl
19353.0 charge
S(6)
2712.0
Use input
N(5)
0.29
gfw
62.0
values; do not
N(-3)
0.03
as
NH4
speciate
O(0)
1.0
O2(g) -0.7
Equilibrate with
Alkalinity
141.682 as HCO3
atmosphere
12
Page 12
Advanced Aqueous Geochemistry
DM Sherman, University of Bristol
2010/2011
Kinetics of Oxidation by O2
• The oxidation of
organic molecules by
triplet oxygen is spinforbidden. Hence
oxidation by triplet
O2 is very slow.
Peroxide, Superoxide and Ozone..
• Peroxide and superoxide are reactive intermediates
resulting from the oxidation of species by oxygen.
• Hydrogen peroxide (H2O2) has O in an oxidation
state = -1.
• Superoxide (O2-1 ) has O in an oxidation state =-1/2.
• The mineral pyrite (FeS2) is a persulfide.
• Ozone (O3) has oxygen in an oxidation state of +2/3.
13
Page 13
Advanced Aqueous Geochemistry
DM Sherman, University of Bristol
2010/2011
Summary
• Understand oxidation-reduction reactions and oxidation
states.
• Nernst Equation.
• Eh-pH diagrams.
• Redox environments in nature.
• Calculate Eh from concentrations.
• Prediction of stable oxidation states.
14
Page 14